- #1
LuculentCabal
- 20
- 0
For a unit mass sliding down the stationary curve [itex]y = f(x)[/itex] at the point [itex](x, y)[/itex], we can consider our mass to be sliding down a stationary inclined plane which is tangential to the curve at the point [itex](x, y)[/itex]. The slope of this inclined plane is thus [itex]\frac{dy}{dx}(x)[/itex].
For the remainder of this thread, [itex]y'[/itex] means [itex]\frac{dy}{dx}[/itex]
For the purposes of this excersize, I want the coefficient of static friction to be zero.
Let [itex]\mu_k[/itex] be the coefficient of kinetic friction
Let [itex]g[/itex] be the acceleration due to gravity
Let [itex]\vec{F_g} = mg[/itex] be the force exerted on our unit mass due to gravity.
Let [itex]\vec{F_{||}}[/itex] be the force exerted on our mass parallel to the plane.
Let [itex]\vec{F_\bot}[/itex] be the force exerted on our mass perpendicular to and into the plane.
Let [itex]\vec{F_N} = - \vec{F_\bot}[/itex] be the force exerted on our mass perpendicular to and out of the plane .
Let [itex]\vec{F_f} = \mu_k \vec{F_\bot}[/itex] be the force of friction which resists [itex]\vec{F_{||}}[/itex]
Then [itex]\vec{F_Ʃ} = \vec{F_{||}} + \vec{F_f} + \vec{F_\bot} + \vec{F_N}[/itex] is the net force exerted on our mass. This is our objective.
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From the vector analysis, I have found [itex]\vec{F_Ʃ}[/itex] to be given by:
[itex]\frac{1}{g} \vec{F_Ʃ} = [cos(θ) sin(θ) - \mu_k cos^2(θ)] \ \hat{i} + [sin^2(θ) - \mu_k cos(θ) sin(θ)] \ \hat{j}[/itex]
Where:
[itex]θ = tan^{-1} (y')[/itex]
The model behaves as expected whenever [itex]\mu_k[/itex] = 0.
However, consider the trivial cases 1. and 2. below:
Case 1.
[itex]\ \ \ g = -9.81[/itex] // arbitrarily
[itex]\ \ \ \mu_k = 0.5[/itex] // arbitrarily
[itex]\ \ \ y = x[/itex] // the standard inclined plane
Case 2.
[itex]\ \ \ g = -9.81[/itex] // same as Case 1.
[itex]\ \ \ \mu_k = 0.5 [/itex] // same as Case 1.
[itex]\ \ \ y = -x[/itex] // this is simply observing Case 1. from the other side of the plane
These parameters result in the following dynamics:
Case 1.
[itex]\ \ \ \frac{d^2 x}{d t^2} = -2.4525[/itex]
[itex]\ \ \ \frac{d^2 y}{d t^2} = -2.4525[/itex]
Case 2.
[itex]\ \ \ \frac{d^2 x}{d t^2} = 7.3575[/itex]
[itex]\ \ \ \frac{d^2 y}{d t^2} = -7.3575[/itex]
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For the previous cases, my intuition told me to expect the following:
a) That the sign on the horizontal acceleration would be negative in Case 1. and positive in Case 2.
b) That the sign on the vertical acceleration would be negative in Case 1. and negative in Case 2.
c) That the magnitude of the horizontal acceleration in Case 1. would be equal to the magnitude of the vertical acceleration in Case 1.
d) That the magnitude of the horizontal acceleration in Case 2. would be equal to the magnitude of the vertical acceleration in Case 2.
e) That the magnitude of horizontal and vertical acceleration in Case 1. would be equal to those of Case 2.
As we can see, the results are in agreement with a), b), c), d). However, hypothesis e) has failed.
Question:
What is the flaw in the model (or hypothesis e) ) that made the results diverge from e) when [itex]\mu_k[/itex] was non-zero (holding all other parameters constant)?
**If it is helpful, I can post my derivation of [itex]\vec{F_Ʃ}[/itex] in a followup.**
For the remainder of this thread, [itex]y'[/itex] means [itex]\frac{dy}{dx}[/itex]
For the purposes of this excersize, I want the coefficient of static friction to be zero.
Let [itex]\mu_k[/itex] be the coefficient of kinetic friction
Let [itex]g[/itex] be the acceleration due to gravity
Let [itex]\vec{F_g} = mg[/itex] be the force exerted on our unit mass due to gravity.
Let [itex]\vec{F_{||}}[/itex] be the force exerted on our mass parallel to the plane.
Let [itex]\vec{F_\bot}[/itex] be the force exerted on our mass perpendicular to and into the plane.
Let [itex]\vec{F_N} = - \vec{F_\bot}[/itex] be the force exerted on our mass perpendicular to and out of the plane .
Let [itex]\vec{F_f} = \mu_k \vec{F_\bot}[/itex] be the force of friction which resists [itex]\vec{F_{||}}[/itex]
Then [itex]\vec{F_Ʃ} = \vec{F_{||}} + \vec{F_f} + \vec{F_\bot} + \vec{F_N}[/itex] is the net force exerted on our mass. This is our objective.
---------------------------------------------------------------------------------------------------------------
From the vector analysis, I have found [itex]\vec{F_Ʃ}[/itex] to be given by:
[itex]\frac{1}{g} \vec{F_Ʃ} = [cos(θ) sin(θ) - \mu_k cos^2(θ)] \ \hat{i} + [sin^2(θ) - \mu_k cos(θ) sin(θ)] \ \hat{j}[/itex]
Where:
[itex]θ = tan^{-1} (y')[/itex]
The model behaves as expected whenever [itex]\mu_k[/itex] = 0.
However, consider the trivial cases 1. and 2. below:
Case 1.
[itex]\ \ \ g = -9.81[/itex] // arbitrarily
[itex]\ \ \ \mu_k = 0.5[/itex] // arbitrarily
[itex]\ \ \ y = x[/itex] // the standard inclined plane
Case 2.
[itex]\ \ \ g = -9.81[/itex] // same as Case 1.
[itex]\ \ \ \mu_k = 0.5 [/itex] // same as Case 1.
[itex]\ \ \ y = -x[/itex] // this is simply observing Case 1. from the other side of the plane
These parameters result in the following dynamics:
Case 1.
[itex]\ \ \ \frac{d^2 x}{d t^2} = -2.4525[/itex]
[itex]\ \ \ \frac{d^2 y}{d t^2} = -2.4525[/itex]
Case 2.
[itex]\ \ \ \frac{d^2 x}{d t^2} = 7.3575[/itex]
[itex]\ \ \ \frac{d^2 y}{d t^2} = -7.3575[/itex]
---------------------------------------------------------------------------------------------------------------
For the previous cases, my intuition told me to expect the following:
a) That the sign on the horizontal acceleration would be negative in Case 1. and positive in Case 2.
b) That the sign on the vertical acceleration would be negative in Case 1. and negative in Case 2.
c) That the magnitude of the horizontal acceleration in Case 1. would be equal to the magnitude of the vertical acceleration in Case 1.
d) That the magnitude of the horizontal acceleration in Case 2. would be equal to the magnitude of the vertical acceleration in Case 2.
e) That the magnitude of horizontal and vertical acceleration in Case 1. would be equal to those of Case 2.
As we can see, the results are in agreement with a), b), c), d). However, hypothesis e) has failed.
Question:
What is the flaw in the model (or hypothesis e) ) that made the results diverge from e) when [itex]\mu_k[/itex] was non-zero (holding all other parameters constant)?
**If it is helpful, I can post my derivation of [itex]\vec{F_Ʃ}[/itex] in a followup.**